That is true because
[tex]\lim_{z\to i}\frac{z^2- 1}{(z+i)(z-i)}[/tex]
does not exist while
[tex]\lim_{z\to i}(z- i)\frac{z^2- 1}{(z+i)(z-i)}= \lim_{z\to i|}\frac{z^2- 1}{z+ i}= i[/tex]
and similarly at z= -i.